Can someone help?

Can someone explain how this comes about. It's a C3 Past Paper in which you have to prove:

tanAsecA + (photo) = sec^2(A)


Posted from TSR Mobile

I'm confused, what exactly are you trying to prove?

Why does that simply to that in the pic?
Then the overall question is below


Posted from TSR Mobile

Oh, multiply the top and bottom by $1-\sin A$. You get $\frac{1-\sin A}{(1+\sin a)(1-\sin A)}$ where the denominator is difference of two squares.

Thanks at least I understand how to get it but it was only a part of a question:

Proves that:
TanASecA +1/1+sinA
Is the same as sec^2(A)


So if I was to do what you done would I not have to multiply the whole line by 1-SinA?


Posted from TSR Mobile

No.

Like: $1 + \frac{2}{3} = 1 + \frac{4}{6}$. We don't need to multiply the whole line by $\frac{2}{2}$. just the fraction.

Anyways: $\frac{1-\sin A}{\cos^2 A} = \frac{1}{\cos^2 A} - \frac{\sin A}{\cos A \cos A} = \sec^2 A - \sec A\tan A$.

Thank you!


Posted from TSR Mobile

No.

Like: $1 + \frac{2}{3} = 1 + \frac{4}{6}$. We don't need to multiply the whole line by $\frac{2}{2}$. just the fraction.

Anyways: $\frac{1-\sin A}{\cos^2 A} = \frac{1}{\cos^2 A} - \frac{\sin A}{\cos A \cos A} = \sec^2 A - \sec A\tan A$.

Thank you!


Posted from TSR Mobile